Half Angle Formulas/Cosine

Theorem

 $\ds \cos \frac \theta 2$ $=$ $\ds +\sqrt {\frac {1 + \cos \theta} 2}$ for $\dfrac \theta 2$ in quadrant $\text I$ or quadrant $\text {IV}$ $\ds \cos \frac \theta 2$ $=$ $\ds -\sqrt {\frac {1 + \cos \theta} 2}$ for $\dfrac \theta 2$ in quadrant $\text {II}$ or quadrant $\text {III}$

where $\cos$ denotes cosine.

Proof 1

 $\ds \cos \theta$ $=$ $\ds 2 \cos^2 \frac \theta 2 - 1$ Double Angle Formula for Cosine: Corollary 1 $\ds \leadsto \ \$ $\ds 2 \cos^2 \frac \theta 2$ $=$ $\ds 1 + \cos \theta$ $\ds \leadsto \ \$ $\ds \cos \frac \theta 2$ $=$ $\ds \pm \sqrt {\frac {1 + \cos \theta} 2}$

We also have that:

In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$
In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$

Proof 2

Define:

$u = \dfrac \theta 2$

Then:

 $\ds \cos^2 u$ $=$ $\ds \frac {1 + \cos 2 u} 2$ Power Reduction Formulas $\ds \leadsto \ \$ $\ds \cos \frac \theta 2$ $=$ $\ds \pm \sqrt {\frac {1 + \cos \theta} 2}$

We also have that:

In quadrant $\text I$, and quadrant $\text {IV}$, $\cos \dfrac \theta 2 > 0$
In quadrant $\text {II}$ and quadrant $\text {III}$, $\cos \dfrac \theta 2 < 0$.

$\blacksquare$